Problem: Find $\lim_{\theta\to \scriptsize\dfrac{\pi}{4}}\dfrac{\cos(2\theta)}{\sqrt 2\cos(\theta)-1}$. Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $\dfrac{1}{2}$ (Choice C) C $\sqrt{2}$ (Choice D) D The limit doesn't exist
Substituting $\theta=\dfrac{\pi}{4}$ into $\dfrac{\cos(2\theta)}{\sqrt 2\cos(\theta)-1}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since our expression includes trigonometric functions, let's try to re-write it using factorization and trigonometric identities. Since we have $\cos(2\theta)$ in our expression, let's rewrite it using one of its double angle identities. Since the expression in the numerator includes $\cos(\theta)$, the most fitting identity is $\cos(2\theta)=2\cos^2(\theta)-1$. $\begin{aligned} &\phantom{=}\dfrac{\cos(2\theta)}{\sqrt 2\cos(\theta)-1} \\\\ &=\dfrac{2\cos^2(\theta)-1}{\sqrt 2\cos(\theta)-1} \gray{\cos(2\theta)\text{ identity}} \\\\ &=\dfrac{(\sqrt 2\cos(\theta)+1)(\sqrt 2\cos(\theta)-1)}{\sqrt 2\cos(\theta)-1} \gray{\text{Diff. of squares}} \\\\ &=\dfrac{(\sqrt 2\cos(\theta)+1)(\cancel{\sqrt 2\cos(\theta)-1)}}{\cancel{\sqrt 2\cos(\theta)-1}} \gray{\text{Cancel common factors}} \\\\ &={\sqrt 2\cos(\theta)+1}\text{, for }\theta\neq \{...,-\dfrac{7}{4}\pi, -\dfrac{1}{4}\pi, \dfrac{1}{4}\pi, \dfrac{7}{4}\pi,...\} \end{aligned}$ This means that the two expressions have the same value for all $\theta $ -values (in their domains) except for $-\dfrac{1}{4}\pi+2\pi k$ or $\dfrac{1}{4}\pi+2\pi k$ for any integer $k$, and specifically $\dfrac{\pi}{4}$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{\cos(2\theta)}{\sqrt 2\cos(\theta)-1}={\sqrt 2\cos(\theta)+1}$ for all $\theta$ -values in the interval $(0,\dfrac{\pi}{2})$ except for $\theta=\dfrac{\pi}{4}$. Therefore, $\lim_{\theta\to \scriptsize\dfrac{\pi}{4}}\dfrac{\cos(2\theta)}{\sqrt 2\cos(\theta)-1}=\lim_{\theta\to \scriptsize\dfrac{\pi}{4}}{\sqrt 2\cos(\theta)+1}=2$ (The last limit was found using direct substitution.) In conclusion, $\lim_{\theta\to \scriptsize\dfrac{\pi}{4}}\dfrac{\cos(2\theta)}{\sqrt 2\cos(\theta)-1}=2$.